How can I compute the following limit, and is there a general method to resolve problems of this type?



marked as duplicate by YuiTo Cheng, Michael Rozenberg calculus Jul 4 at 5:13

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  • $\begingroup$ $(+\infty)^2 =+\infty$ $\endgroup$ – Paolo Leonetti Oct 10 '16 at 14:48
  • $\begingroup$ I think inside need to be $x[(1+1/x)^x-e]$ $\endgroup$ – student forever Oct 10 '16 at 14:50
  • $\begingroup$ I agree, but we cannot also try to interpret what he wants to ask $\endgroup$ – Paolo Leonetti Oct 10 '16 at 14:52
  • 1
    $\begingroup$ @studentforever You may be right as $$(x+\frac{1}{x})^x=e^{x(log(x(1+\frac{1}{x^é})))}=e^{x(log(x)+log((1+\frac{1}{x^2}))))}$$ tends to infinity. Wait for the OP reaction. $\endgroup$ – Duchamp Gérard H. E. Oct 10 '16 at 14:53
  • $\begingroup$ Yeah fixed the typo... $\endgroup$ – Joshua Benabou Oct 10 '16 at 15:04

Using the Taylor Series for $\log(1+x)$, we get $$ \begin{align} x\log\left(1+\frac1x\right) &=x\left(\frac1x-\frac1{2x^2}+O\left(\frac1{x^3}\right)\right)\\ &=1-\frac1{2x}+O\left(\frac1{x^2}\right) \end{align} $$ Therefore, using the Taylor Series for $e^x$, we get $$ \begin{align} x\left[\left(1+\frac1x\right)^x-e\right] &=x\left[e^{1-\frac1{2x}+O\left(\frac1{x^2}\right)}-e\right]\\ &=xe\left[e^{-\frac1{2x}+O\left(\frac1{x^2}\right)}-1\right]\\ &=xe\left[-\frac1{2x}+O\left(\frac1{x^2}\right)\right]\\ &=-\frac e2+O\left(\frac1x\right) \end{align} $$ Thus, $$ \lim_{x\to\infty}x\left[\left(1+\frac1x\right)^x-e\right]=-\frac e2 $$

  • $\begingroup$ Very nice reasoning +1 $\endgroup$ – Ahmad Bazzi Oct 10 '16 at 15:55

hint:$$\lim_{x\to\infty}{x\left(\left(1+\frac{1}{x}\right)^x-e\right)}=\lim_{x\to\infty}\frac{{\left(1+\frac{1}{x}\right)^x-e}}{\frac{1}{x}}=$$ then apply L'Hospital rule

  • $\begingroup$ L'Hospital seems to give something pretty complicated. Perhaps a second application with some simplification might help. $\endgroup$ – robjohn Oct 10 '16 at 16:11

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